# Probability of one pair of brothers sitting together

I have 10 kids which are 5 pairs of brothers sitting randomly across 5 tables, each table has two seats. What is the probability of only one couple sitting together?

Solving it with combinatorics alone (as it was done in this answer) is complicated because After pairing the first couple I have too many options. So I was trying to think of the complement event which is only $$\ 1$$ couple of brothers not sitting together, which is zero (because it can not be that only 1 couple are not sitting together). So what is the best way to solve it?

• The complement is "not only one couple together " and differs from "only one couple not together ". – drhab Dec 9 '18 at 8:09
• I don't think I understand.. – bm1125 Dec 9 '18 at 8:13
• You mean to say $5$ pairs of brothers and sisters? Is the order in which a pair sits on a given table significant? – Shubham Johri Dec 9 '18 at 8:35
• Yes lets say brothers and sisters. the order in which a pair sits is insignificant – bm1125 Dec 9 '18 at 8:46
• The complement of "only one couple together" is in this context: "a number of couples together that equals $0,2,3,4$ or $5$" (all numbers in $\{0,1,2,3,4,5\}$ except $1$). That is something else than "only one couple not together" or equivalently "exactly $4$ couples together". – drhab Dec 9 '18 at 14:25

I have $$8$$ kids which are $$4$$ pairs of brothers sitting randomly across $$4$$ tables, each table has two seats. What is the probability that no couple sits together?

Number the couples $$1,2,3,4$$ and let $$B_{i}$$ denote the event that couple $$i$$ sits together.

Then to be found is $$P\left(B_{1}^{\complement}\cap B_{2}^{\complement}\cap B_{3}^{\complement}\cap B_{4}^{\complement}\right)=1-P\left(B_{1}\cup B_{2}\cup B_{3}\cup B_{3}\right)$$

Applying the principle of inclusion/exclusion and also symmetry we find that this probability equals:

$$1-4P\left(B_{1}\right)+6P\left(B_{1}\cap B_{2}\right)-4P\left(B_{1}\cap B_{2}\cap B_{3}\right)+P\left(B_{1}\cap B_{2}\cap B_{3}\cap B_{4}\right)$$

Here we find:

• $$P\left(B_{1}\right)=\frac{1}{7}$$
• $$P\left(B_{1}\cap B_{2}\right)=P\left(B_{2}\mid B_{1}\right)P\left(B_{2}\right)=\frac{1}{5}\frac{1}{7}$$
• $$P\left(B_{1}\cap B_{2}\cap B_{3}\right)=P\left(B_{3}\mid B_{1}\cap B_{2}\right)P\left(B_{1}\cap B_{2}\right)=\frac{1}{3}\frac{1}{5}\frac{1}{7}$$
• $$P\left(B_{1}\cap B_{2}\cap B_{3}\cap B_{4}\right)=P\left(B_{4}\mid B_{1}\cap B_{2}\cap B_{3}\right)P\left(B_{1}\cap B_{2}\cap B_{3}\right)=\frac{1}{1}\frac{1}{3}\frac{1}{5}\frac{1}{7}$$

So we find: $$P\left(B_{1}^{\complement}\cap B_{2}^{\complement}\cap B_{3}^{\complement}\cap B_{4}^{\complement}\right)=1-4\cdot\frac{1}{7}+6\cdot\frac{1}{5}\frac{1}{7}-4\cdot\frac{1}{3}\frac{1}{5}\frac{1}{7}+\frac{1}{1}\frac{1}{3}\frac{1}{5}\frac{1}{7}=\frac{4}{7}$$

Now we step to the original problem.

Again number the couples $$1,2,3,4,5$$ and let $$E_{i}$$ denote the event that couple $$i$$ will sit together.

Further let $$E$$ denote the event that exactly one couple sits together.

Then $$E$$ is the union of the events:

• $$E_{1}\cap E_{2}^{\complement}\cap E_{3}^{\complement}\cap E_{4}^{\complement}\cap E_{5}^{\complement}$$
• $$E_{1}^{\complement}\cap E_{2}\cap E_{3}^{\complement}\cap E_{4}^{\complement}\cap E_{5}^{\complement}$$
• $$E_{1}^{\complement}\cap E_{2}^{\complement}\cap E_{3}\cap E_{4}^{\complement}\cap E_{5}^{\complement}$$
• $$E_{1}^{\complement}\cap E_{2}^{\complement}\cap E_{3}^{\complement}\cap E_{4}\cap E_{5}^{\complement}$$
• $$E_{1}^{\complement}\cap E_{2}^{\complement}\cap E_{3}^{\complement}\cap E_{4}^{\complement}\cap E_{5}$$.

These events are mutually excusive and equiprobable so that:

$$P\left(E\right)=5P\left(E_{1}^{\complement}\cap E_{2}^{\complement}\cap E_{3}^{\complement}\cap E_{4}^{\complement}\cap E_{5}\right)=5P\left(E_{1}^{\complement}\cap E_{2}^{\complement}\cap E_{3}^{\complement}\cap E_{4}^{\complement}\mid E_{5}\right)P\left(E_{5}\right)$$

Here $$P\left(E_{5}\right)=\frac{1}{9}$$ because - after placing one brother on a chair - there are $$9$$ chairs left and only $$1$$ of them results in couple $$1$$ at the same table by placing the other brother.

Working under condition $$E_{5}$$ there are $$4$$ tables left for the remaining $$4$$ couples and we are back in the problem that was solved first.

So $$P\left(E_{1}^{\complement}\cap E_{2}^{\complement}\cap E_{3}^{\complement}\cap E_{4}^{\complement}\mid E_{5}\right)$$ equals the $$P\left(B_{1}^{\complement}\cap B_{2}^{\complement}\cap B_{3}^{\complement}\cap B_{4}^{\complement}\right)$$ that was calculated there and we end up with:

$$P\left(E\right)=P\left(E_{1}^{\complement}\cap E_{2}^{\complement}\cap E_{3}^{\complement}\cap E_{4}^{\complement}\mid E_{5}\right)P\left(E_{5}\right)=\frac{4}{7}\frac{1}{9}=\frac{4}{63}$$

• Thank you! this is really informative answer. I'm trying to apply what you taught me here to a case of 2 couples sitting together. So if $\ E$ is the event of only two couples sitting together, it is true that $\ P(E) = 45 \cdot P(E_1 \cap E_2 \cap E_3^c \cap E_4^c \cap E_5^c)$ because they are excursive and equiprobable so $P(E) = 45 P(E_3^c \cap E_4^c \cap E_5^c | E_1 \cap E_2)P(E_1 \cap E_2) = \frac{1}{3} \cdot \frac{1}{9} \cdot \frac{1}{7}$ ? – bm1125 Dec 9 '18 at 15:28
• Not factor $45$ but factor $\binom52=10$. So $P(E)=10P(E_3^c\cap E_4^c\cap E_5^c\mid E_1\cap E_2)P(E_1\cap E_2)$. This with $P(E_1\cap E_2)=P(E_1)P(E_2\mid E_1)=\frac19\frac17$. To find the conditional probability (I haven't done that yet) you use the same strategy as above: find the probability that by $3$ couples and $3$ tables no couple will sit together. – drhab Dec 9 '18 at 15:40